Order, degree, and solutions

Before solving a differential equation we must classify it. The two basic descriptors are the order (the highest derivative present) and the degree (the power of the highest-order derivative after rationalisation). Together they tell us what kind of solution to expect and which technique to bring.

Definitions

A differential equation is an equation that relates a function $y$ of an independent variable $x$ with one or more of its derivatives $\dfrac{dy}{dx}, \dfrac{d^2y}{dx^2}, \ldots$.

Order: the highest-order derivative present.

Degree: the power of the highest-order derivative, after the equation is rationalised , i.e. cleared of fractional and surd expressions involving derivatives.

For example, $\dfrac{d^2 y}{dx^2} + 3\dfrac{dy}{dx} - 5y = 0$ has order $2$, degree $1$.

The equation $\left(\dfrac{dy}{dx}\right)^3 + \sin y = 0$ has order $1$, degree $3$.

The equation $\sqrt{\dfrac{dy}{dx}} = x$ has order $1$. Degree: squaring both sides gives $\dfrac{dy}{dx} = x^2$, so degree $1$.

The equation $\sqrt[3]{\dfrac{d^2 y}{dx^2}} = \dfrac{dy}{dx} + 1$ becomes (after cubing) $\dfrac{d^2 y}{dx^2} = \left(\dfrac{dy}{dx} + 1\right)^3$. Order $2$, degree $1$.

If after rationalisation the highest-order derivative cannot be brought to a polynomial form (e.g. $\sin(y'') + y = 0$), the degree is not defined.

What is a solution?

A solution to a differential equation on an interval $I$ is a function $y = \phi(x)$ defined on $I$ such that substituting $y = \phi(x)$ (and its derivatives) into the equation makes it true for every $x \in I$.

For example, $y = e^x$ is a solution of $\dfrac{dy}{dx} = y$ on all of $\mathbb{R}$. So is $y = 2e^x$. So is $y = Ce^x$ for any constant $C$. The full family $y = Ce^x$ is the general solution. Choosing a specific $C$ (typically via an initial condition like $y(0) = 3$, which forces $C = 3$) gives a particular solution.

The general principle: a differential equation of order $n$ has a general solution with $n$ arbitrary constants. To pin down a particular solution we need $n$ pieces of additional information.

Forming a DE from a family of curves

The reverse problem also matters. Given a family of curves with $n$ arbitrary constants , e.g. $y = a x + b$ (two constants), or $y = A \sin(x + \alpha)$ , find a differential equation, of order $n$, that the entire family satisfies.

Procedure. Differentiate the equation $n$ times. You now have the original equation plus $n$ derivative equations, in total $n + 1$ equations involving $x$, $y$, $y'$, …, $y^{(n)}$, and the $n$ constants. Eliminate all $n$ constants , the result is a differential equation of order $n$.

Example. Form the DE for $y = a x + b$.

Differentiate: $y' = a$. Differentiate again: $y'' = 0$. The second equation contains no constants , it is the DE we want. So the DE is $\dfrac{d^2 y}{dx^2} = 0$.

Example. Form the DE for $y = Ae^{2x} + B e^{-2x}$.

$y' = 2A e^{2x} - 2B e^{-2x}$. $y'' = 4A e^{2x} + 4B e^{-2x} = 4y$. So the DE is $\dfrac{d^2 y}{dx^2} - 4y = 0$.

Verifying a solution

You can always test a proposed solution by substitution. Given $y = \phi(x)$, compute the derivatives, plug into the equation, and check that it simplifies to $0 = 0$ (or the appropriate identity).

Worked examples

Example 1. Find the order and degree of $\left(\dfrac{d^2 y}{dx^2}\right)^2 + \dfrac{dy}{dx} - 5 = 0$.

Highest derivative: second. Order $2$. After rationalisation: already polynomial in $y''$. Degree $2$.

Example 2. Find the order and degree of $\sqrt{1 + \left(\dfrac{dy}{dx}\right)^2} = \dfrac{d^2 y}{dx^2}$.

Square both sides: $1 + (y')^2 = (y'')^2$. Order $2$, degree $2$.

Example 3. Form the DE for the family $y = mx$, where $m$ is an arbitrary constant.

Differentiate: $y' = m$. Substitute back: $y = y' \cdot x$, i.e. $x \dfrac{dy}{dx} - y = 0$.

Example 4. Verify that $y = a \cos x + b\sin x$ satisfies $\dfrac{d^2 y}{dx^2} + y = 0$.

$y' = -a \sin x + b \cos x$, $y'' = -a \cos x - b \sin x = -y$. So $y'' + y = 0$. Confirmed.

Example 5. Find the DE of all circles passing through the origin with centres on the $x$-axis.

Family: $(x - a)^2 + y^2 = a^2$, expanding to $x^2 - 2ax + y^2 = 0$. One constant $a$, so order $1$ expected. Differentiate: $2x - 2a + 2yy' = 0$, giving $a = x + yy'$. Substitute into the original: $x^2 - 2(x + yy')x + y^2 = 0$, simplifying to $-x^2 - 2xyy' + y^2 = 0$, or $y^2 - x^2 = 2xy \dfrac{dy}{dx}.$

Example 6. Find the DE of the family of ellipses $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, considering $b$ a function of $a$, both parameters arbitrary. (Two constants, so order $2$.)

Differentiate: $\dfrac{2x}{a^2} + \dfrac{2yy'}{b^2} = 0$, so $\dfrac{x}{a^2} = -\dfrac{yy'}{b^2}$, hence $\dfrac{x}{y y'} = -\dfrac{a^2}{b^2}$. Differentiate again to eliminate the second constant: $\dfrac{(yy')(1) - x(y'^2 + y y'')}{(yy')^2} = 0$, so $yy' = x(y'^2 + yy'')$, i.e. $xyy'' + x(y')^2 - yy' = 0$.

Try it yourself

1. Find the order and degree: $y' + 5y = \cos x$.
2. Find the order and degree: $(y'')^3 + 4 y' - 2 = 0$.
3. Find the order and degree: $\dfrac{d^3 y}{dx^3} + y^3 = 0$.
4. Find the order and degree: $\left(\dfrac{dy}{dx}\right)^{2/3} = x$.
5. Form the DE for $y = c x^2$, $c$ arbitrary.
6. Form the DE for $y = a \sin(x + b)$, $a$, $b$ arbitrary.
7. Form the DE for $y = c_1 e^x + c_2 e^{2x}$.
8. Verify $y = \sin x$ satisfies $y'' + y = 0$.
9. Verify $y = e^{2x}$ satisfies $y'' - 4y = 0$.
10. Form the DE of the family $x^2 + y^2 = r^2$ (circles centred at origin).
11. Find a particular solution of $y' = 2x$ satisfying $y(1) = 3$.
12. Verify $y = x \log x$ satisfies $x y' - y = x^2$.
13. Form the DE for the family $y = A e^{3x} + B$.
14. State why $y' = |y'| + 1$ has no real-valued solution.

Pitfalls and tricks

• Order is always defined; degree sometimes isn't. When the equation contains a transcendental function of derivatives (like $\sin y'$), the degree is undefined.
• Rationalise before reading degree. Always clear fractions, radicals, and exponents involving derivatives.
• Number of constants = order. When forming a DE, the order equals the number of arbitrary constants in the family.
• Verify by substitution. Mathematics is the only subject where you can check your own answer cold.
• Eliminate constants by sequential differentiation. Each derivative reduces the problem by one constant.